Question

An electron is released with a velocity of $5\times 10^{6}m{s}^{-1}$ in an electric field of $10^{3}N{C}^{-1}$ which has been applied so as to oppose its motion. How much time could it take before it is brought to rest?

• A. $2.8\times 10^{-8}s$
• B. $1.8\times 10^{-8}s$
• C. $6.4\times 10^{-8}s$
• D. $4.2\times 10^{-8}s$

Answer 1

Answer: (A)

Since electric field is applied so as to oppose the motion of electron,
$a=-\frac{eE}{m}$ (retardation)
$=-\frac{1.6\times 10^{-19}\times 10^3}{9.1\times 10^{-31}}=-1.758\times 10^{14}m{s}^{-2}$
Now, $u=5\times 10^{6}m{s}^{-1},v=0$
Using the relation: $v^2-u^2=2as$, we get
$s=7.11\times 10^{-2}m$
Using the relation: $v=u+at$, we get
$t=2.844\times 10^{-8}s$